Jorge's subset products

First, we can factor $2013=3 * 11 * 61$ . As the product of the subset must be divisible by 2013, it will need to contain at least one multiple of each factor. As $2013=61 * 11 * 3=61 * 33$ , we know that 61 has 33 multiples between 1 and 2013. Subtracting, we find that there are $2013-33=1980$ numbers that aren't multiples of 61. This gives us a list of 1980 numbers which do not satisfy the conditions.

Using the pigeonhole principle, we know that any subset with 1981 will have at least one multiple of 61. We also know that this generic subset of 1981 elements must have factor(s) of 3 and 11 as both numbers have more factors than 61. Finally, as the problem asks for the last three digits, we get 981.

2013 =3 11 61

So in each chosen set must have at least one triplet of the multiples of {3,11,61}.

As the number of multiples of 61 from 1 to 2013(2013/61 = 33) is less than either 3 or 11 ,we can just work with 61.From 1 to 2013 there are 2013 - 33 = 1981 numbers which are not a multiple of 61 . Thus any set of (1980 +1 = 1981) numbers from S would definitely have a triplet of multiples of 3,11 and 61.

Thus n=1981,last three digits of n=981.

3,11,and 61 is the prime factor of 2013. the largest is 61. in 2013, there are 33 number divisible by 61 but we just need 1 so 2013 - 32 = 1981. we ask to write 981.

The prime factors of 2013 are 3, 11 and 61. So our set of n elements must contain some multiple of each of these numbers in order for the product of the elements to be divisible by 2013.

The set S contains (2013 / 3) = 671 multiples of 3, (2013 / 11) = 183 multiples of 11 and (2013 / 61) = 33 multiples of 61. Since 1980 elements of S are not multiples of 61, it then follows that in order to force any choice of n distinct elements to include all of these, n must be at least 1981.

$2013$ = $3 \times 11 \times 61$

There are - list

1342 integers in the set that are not divisible by 3 1830 integers in the set that are not divisible by 11 1980 integers in the set that are not divisible by 61

According to the Pigeon Hole Principle, there will be multiples of 3, 11, 61 if we pick at least 1981 integers from the given set. Thus, n = 1981.

Divisors of $2013 = 1, 3, 11, 33, 61, 183, 671, 2013$

First we add the number of numbers between $1$ and $2013$ which aren't divisible by 3, 11, 61

Multiples of 3 ⇒671

Multiples of 11 ⇒183

Multiples of 61 ⇒33

Multiples of 3 and 11 ⇒61

Multiples of 3 and 61 ⇒11

Multiples of 11 and 61 ⇒3

Multiples of 3, 11, and 61 ⇒1

$671 + 183 + 33 - 61 - 11 - 3 + 1 = 813$

$2013 - 813 = 1200$

A multiple of 2013 is a multiplication of 3s, 11s, and 61s. So add the result above with the number of the multiples of 3 and 11 only, then add 1 from the multiple of 61.

$671 + 183 - 61 - 11 - 3 + 1 = 780$

$1200 + 780 + 1 = 1981$

$n = 1981$

The last three digits of $n$ = 981

The prime factorization of 2013 is 3 11 61. Of these 3 factors, 61 has the least multiples from 0-2013. Therefore, by some pigeonholing, the value of n will be determined by the number of possible multiples of 61 in the set. 2013/61=33. Thus, n = 2013-32 = 1981. The last 3 digits of this n are 981.